Answer:
1.67 m/s
Explanation:
Since this is an inelastic collision (the block and the arrow stick together after the collision), we can solve the problem by using the law of conservation of momentum:
[tex]p_i = p_f\\m_a v_a + m_b v_b = (m_a + m_b) v[/tex]
where
[tex]m_a = 2.0 kg[/tex] is the mass of the arrow
[tex]v_a = 10 m/s[/tex] is the initial speed of the arrow
[tex]m_b = 10 kg[/tex] is the mass of the block
[tex]v_b = 0[/tex] is the initial speed of the block
[tex]v = ?[/tex] is the final speed of the arrow+block
Substituting and re-arranging the equation, we find:
[tex]v=\frac{m_a v_a + m_b v_b}{m_a +m_b}=\frac{(2.0 kg)(10 m/s)+0}{2.0 kg+10.0 kg}=1.67 m/s[/tex]
